A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
If y = 5/2(e^(x/5) + e^(x/5)), prove that y'' = y/25.
anonymous
 5 years ago
If y = 5/2(e^(x/5) + e^(x/5)), prove that y'' = y/25.

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I expanded the bracket: y = 5/2(e^(x/5)) + 5/2(e^(x/5))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then I took the first derivative:

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y ' = 5x/10(e^(x/5))  5x/10(e^(x/5))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now how should I go about with getting the second derivative? Should I factor out the 5x/10 and then use product rule?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y'=5/2[e ^{x/5}(1/5)+e^{x/5}(1/5)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y'=(5/2)(1/5)[e ^{x/5}e ^{x/5}]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does it make a difference if we were to expand first with 5/2, and then take the first derivaitve?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because that is what I did and then I took the dy/dx of each of the two terms.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but y did u multiply with x/5?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0instead it is 1/5 the coefficient of x/5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No sorry I did the following: y = 5/2(e^{x/5}) + 5/2(e^{x/5}) y '= 5/2(1/5)*(e^{x/5}) + 5/2(1/5)(e^{x/5})

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So then: y = 1/2(e^{x/5} + e^{x/5})

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y' = 1/2(e^{x/5} + e^{x/5})

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0n there must b negative between to terms

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0coz 1/5 is the common factor

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y' = 1/2(e^{x/5}  e^{x/5}), right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So then for y'' I get..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y''=1/2[e ^{^{x/5}}(1/5)e ^{x/5}(1/5)]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes I got that, then I expanded, and I got: y'' = 1/10(e^{x/5} + e^{x/5})

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y''=1/10[e ^{x/5}+e ^{x/5}]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0again no need to expand

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[e ^{x/5}+ e ^{x/5}=2y/5\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How did you get e^x/5 + e^x/5 = 2y/5?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got it thank you so much. So when you are taking derivative of a function that has a constant which is multipled by a bracket that can be differentiated, you just take the derivative of the terms inside the bracket and multiply by the constant in front of it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok awesome. Thank you for your help.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.